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Open differential:Force at the planetary wheel

  1. Mar 30, 2014 #1
    http://imageshack.com/a/img33/5853/ge7y.gif [Broken]

    In the above diagram,the part 3 the planetary wheel(its not the gear at the input or output side but in the middle of the differential). I call it the planetary wheel because I consider a bevel gear differential just an approximation of a planteary gear differential with equal radii of ring and sun gears. First of all,I want to clarify:

    1.A open differential transfers equal power to the left and right wheels,right? P_left=P_right
    I make this presumption because when taking a turn,the angular velocities of the wheel are different and hence to make the power equal,I would need to vary the torques.I mean:
    M_left * ω_left =M_right *ω_right

    So,if
    ω_left<ω_right => M_left > M_right

    Am I right with this logic?

    2. If the above logic is correct, in the above diagram the force [F_E /2] would vary at the left and right wheels. If [F_E/2] were equal,the planetary gears would revolve and not rotate around their axis(straight line driving). But,in case of the turns, what would the force magnitude at the planetary wheel be(marked with a question mark in above diagr.)? I presume this would be [F_left-F_right]. Am I correct?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 30, 2014 #2

    jack action

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    Science Advisor
    Gold Member

    No, P_left ≠ P_right.

    But T_left = T_right, no matter what.

    From http://zhome.com/ZCMnL/tech/Torsen/Torsen.htm:
    Thought experiment: It is possible to have one wheel rotating on an ice patch while the other is not rotating because it is on asphalt. If one wheel has 0 rpm, then it must have no power (P = Tω), hence -obviously - it cannot have the same power as the other wheel.
     
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